Determining the percentage composition of Vinegar.

Question

In this lab, we were titrating CH₃COOH with NaOH. we used an average of 33.433 mL of 0.7 molarity NaOH to neutralize 25mL of CH₃COOH.

The questions are:

calculate the moles of NaOH from the average volume used.

Calculate the moles of acetic acid present originally.

Calculate the molarity of the acetic acid in the solution.

Calculate the mass of acetic acid in 1.00 L of solution.

Calculate the percentage of the acetic acid in the vinegar.

If you can answer any of these it would really help me in finishing the lab report with how you've done it so I can understand it. Thank you for the help

Answer 1

This is very similar to your previous question using HCl and NaOH. The main difference is here you are using acetic acid (CH₃COOH) instead of hydrochloric acid (HCl). The base (NaOH) remains the same.

Again, you should write the balanced equation so you can see what's going on:

CH₃COOH + NaOH ==> CH₃COONa + H₂O ... balanced equation for neutralization reaction

As with your previous question, I'm going to round the volume of NaOH to 33.4 mls as I doubt you measured it any accurately than that, and 0.7 M NaOH has only 1 significant figure anyway. So, here we go...

Find the mols of NaOH used from the volume and the molarity:

33.4 mls NaOH x 1 L / 1000 mls x 0.7 mol / L = 0.0234 moles

Find moles of acetic acid present originally:

0.0234 mols NaOH x 1 mol CH₃COOH / mol NaOH = 0.0234 mols CH₃COOH

Find molarity of acetic acid:

0.0234 mols CH₃COOH / 25 mls = 0.0234 mols / 0.025 L = 0.936 M = 0.9 M (1 sig. fig.)

Find the mass of CH₃COOH in 1.00 L:

From the molarity (0.9 M) we know there are 0.936 mols acetic acid in 1 liter. Molar mass of CH₃COOH = 60.05 g/mol. So, 0.936 mols / L x 60.05 g / mol = 56.2 g acetic acid in 1 liter.

Find the % acetic acid in the sample of vinegar:

The answer to this part depends on if you want the % m/m, v/v or m/v. We know the mass of acetic acid (56.2 g) and we know the volume (1 L or 1000 mls), so the only basis we can use is % m/v. That is calculated as

56.2 g / 1000 ml (x100%) = 5.62% (mass/volume)